text:
Why should the leading edge of the horizontal tail be lower than the trailing edge?
As mentioned in this answer: https://drones.stackexchange.com/a/2564/5088

The leading edge of the tail should be lower than its trailing edge, providing some "down force" in flight.

I am confused why a down force is required, because I thought that it would not produce lift. Can somebody explain why the downforce is required?

Comments are not for extended discussion; this conversation has been moved to chat.

In conventional aircraft design, the tailplane usually provides a downforce to insure longitudinal stability around the pitch axis, which is also the Center of Gravity.

Photo Source
The center of gravity is forward of the main lifting service and the tail is well back providing a downward force.
In normal normal cruise, the two forces balance each other. If turbulence or some other force causes the nose to rise, the aircraft will start to slow, and the angle of attack will increase on the lifting wing, but decrease on the downward acting tai. This causes the less lift on the wing but also less downward force on the tailplane. This will then cause the nose to lower and return back to the cruise attitude.
If the nose should drop, the speed will increase, and the angle of attack will decrease on the wing, but increase on the downward acting tail. This causes more downward force on the tailplane, which will then cause the nose to raise back to the normal cruise attitude. This is called positive stability.
As you load the aircraft with a more rearward C of G, less downward force is required, and there will be less induced drag from the rear tailplane. This can give you less fuel consumption and better range.
If you move the C of G too far rearward, the tail will need to be trimmed to provide much less downforce, or even a upwards force. In this situation you will have negative stability because changes in speed and angle of attack will now have the opposite of the desired effect.  Any decrease in speed and increase in angle of attack will cause the the nose to rise further, and any increase in speed and decrease in angle of attack, will cause the nose to drop further.
A rear C of G outside the normal limits can make the aircraft difficult to control or make recovery from a stall impossible.

Please do not leave the impression that positive tail lift equals negative stability. This nonsense has been repeated much too often.

In most conventional aircraft, the C of G is forward of the center of lift and the tail must provide downforce. An aircraft loaded with a C of G outside it’s rear limit will need positive lift, and it will have negative stability. Why is this nonsense?

For you homebuilders out there, the tail trims the wing, the wing does the lifting.  Be aware that wing center of pressure can move as AoA changes.  Design tail accordingly. More info here.

A good design will keep tail load small, negative at low AoA and positive at high AoA. This leaves the largest margin for control inputs. And positive tail lift can easily be combined with allowed cg limits - only at a forward cg position will the tail lift stay negative through the whole AoA range.

Please modify your answer to make it clear that movement of the nose relative to the horizon, and a change in airspeed, are NOT the factors causing the instability. It is the change in Angle of Attack (AOA). If overall aircraft AOA momentarily increases (gust of wind  or whatever), then the main wing AOA increases, increasing lift, and nose down moment, but the horizontal tail AOA (negative) is reduced ( less negative), reducing down force and reducing nose UP moment. Result? The aircraft overall moment is nose down, reducing AOA to where the moments are again in balance.

That is the essence of positive longitudinal stability.  And it all happens as I described, as AOA changes, no matter whether the nose is up, down or sideways, and no matter whether the airspeed is increasing, decreasing, or stable.

@Charles Bretana Thanks for the added info. My knowledge of aerodynamics is mostly self-taught and it makes sense to me now that AOA is a much bigger factor than speed.

Not only a bigger factor, the only factor.

Airplanes with conventional tails use the tail surface to "trim" the wing, to force it to operate at an angle to the airflow so that it produces lift, as well as provide stability via "weathervane effect".
Think of a weathervane on a barn, which points into wind. It's statically stable. I need to make it point into wind at an offset angle for some reason. So I'll "trim" the vane by bending the trailing edge over. As the wind shifts, the vane will point into it, but with some constant offset angle because of the trimming force applied by the bent trailing edge. If I bend the trailing edge more, it will point with an increasing offset angle. I could call this the vane's "angle of attack".
In an airplane I take this effect and rotate the axis 90 degrees, and because it's now a free body in space, the pivot axis becomes the center of mass of the entire thing. Absent any trimming force, my airplane merely becomes a very stable ballistic object, like a lawn dart.
The plane has static stability, it wants to point into wind, because the "pivot", the C of G, is ahead of the center of all of the aerodynamic forces acting vertically, including the fuselage, engine nacelles, and tail. Like the weathervane on a barn turned sideways, but as a free body in space, more of a lawn dart.
To make the "weathervane" actually fly along, resisting gravity to hold the thing up instead of arcing into the ground as a very stable lawn dart, you have to "trim" it to force it to align itself into the airflow with a positive angle on the wings. If you want the wings to lift up, you have to force it to operate the wings at a positive angle, and to make it operate at a positive angle, you have to apply a trimming force in the opposite direction back at the tail, like bending the trailing edge of the weathervane.
Which means, if you want the wings to lift up, the tail has to be generating a trimming force downward, and so the tail's local angle of attack will be negative in order to force the main wing to be aligned into the airflow at a positive angle. This is what it means by the reference to having the leading edge of the tail lower than the wing. This opposing force balance is needed to make the wing operate at an angle to the flow that holds everything up.
For the center of mass, as the effective pivot axis, it just has to be ahead the aerodynamic center of the whole thing, like a weathervane, to be statically stable. Center of mass position, if too far aft, kills the weathervaning effect (stability is neutral), or if far enough aft, flips the weathervaning effect around the wrong way (stability is negative).
So your model has to have two things to fly properly.  If you turn it sideways and attach it to a pivot of some sort, and you locate the pivot at its center of mass, it needs to point into wind, whatever direction the wind is coming from. That confirms positive static stability. Then you need to configure the tail to "trim" it, so that when it aligns into the wind, it aligns with an offset so that the air hits the main wing at a positive angle. It will now achieve its stable state with the wing at a positive angle to the airflow. If you then hold it horizontally and throw it, it should fly, more or less.

Short Answer: This is a result of making the aircraft stable in longitudinal axis.

Longitudinal Stability Basics

Nomenclature: $C_m$ is the  pitching moment coefficient. Pitch-up moment is denoted by +ve $C_m$. Also, $α$ denotes Angle of Attack.

If the AC of a surface is ahead of CG, then that surface has a destabilizing contribution. This can be understood through this image:

A positive $∆α$ will produce a positive $∆C_m$ (pitch-up). This becomes a vicious cycle, since pitching-up will further increase the $α$ - that in turn will increase the $C_m$ further (greater pitch-up moment) and so on. Basically, the aerofoil will have a tendency to diverge away from equilibrium - on its own, it is statically unstable.
Likewise, if AC is behind CG, a positive $∆α$ produces a negative $∆C_m$ (pitch-down).  Pitching down will reduce the $α$ - the aerofoil has a tendency to return back to its trim $α$ and is thus statically stable.
Making a stable aircraft
A conventional aircraft primarily has two surfaces that determine its longitudinal stability - the wing and the tail.²

Wing: Its contribution can either be stabilizing or destabilizing, depending on the position of CG relative to its AC.

Tail: Its contribution is always stabilizing, since CG is always ahead of tail AC.

Consider an aircraft with its wing AC ahead of aircraft CG. In that case, the wing has a destabilizing contribution.
Wing contribution: (destabilizing) $$\frac{dC_{m\ wing}}{dα} > 0$$
Tail contribution: (always stabilizing) $$\frac{dC_{m\ tail}}{dα} < 0$$
It is desirable to have a wing on an aeroplane, but a conventional wing on its own is unstable. To fix this, we add a tail.
Now, how do we ensure that the tail would be enough? It is necessary to ensure that the stabilizing contribution of tail is greater than the destabilizing contribution of wing:
$$ \left( \frac{dC_{m\ wing} + dC_{m\ tail}}{dα} \right) < 0 $$
This ensures that the aircraft as a whole is statically stable. But how is this achieved in practice? - By putting the CG ahead of the neutral point
CG position
Neutral point: This is the point on the longitudinal axis at which positioning the CG results in the aircraft having no tendency:

to return to the equilibrium (stability) and,

to diverge away from the equilibrium (instability).

If an aircraft with neutral stability is subjected to a disturbance that changes its $α$, then that will produce zero net $C_m$. This means that the aircraft will find itself in a new equilibrium, and will now maintain this new $α$.
At neutral point, wing's destabilizing moment is equal in magnitude to tail's stabilizing moment - that's why their combination is neutral.
If CG is moved ahead of the neutral point, the aircraft becomes stable. This is because:

Tail moment arm increases: This increases its stabilizing effect.

Wing moment arm reduces: In our example where wing AC is ahead of CG, wing arm reduces. This reduces its destabilizing effect.

Likewise, if CG is moved aft of the neutral point, the aircraft becomes unstable. This is because wing's unstable contribution increases and tail's stable contribution decreases.
Till now, we have considered an example where wing AC is ahead of CG. In practice, the stability required is usually so large that it can only be achieved by moving CG well ahead of wing AC. This is the configuration we see in most conventional aircraft.
Trimming the moments
With the aircraft in equilibrium at the neutral point:

the positive $C_{m\ wing}$ will be equal in magnitude to the negative $C_{m\ tail}$.

Also, a change in $α$ will produce proportional change in the $C_{m}$ of both the wing and the tail, such that no net change in pitching moment is produced.

Moving CG ahead of the NP reduces +ve wing moment and increases -ve tail moment. The combined result is a net -ve $C_{m}$. To trim that out, tail incidence must be reduced. This produces what is known as "longitudinal dihedral" or "decalage" or whatever you wish to call it.

Dihedral: Angle between two planes.

Longitudinal dihedral: Angle between wing and tail. More specifically, the difference between wing incidence and tail incidence. A positive LD is where wing incidence is greater than tail incidence.

For a longitudinally stable aircraft, trim will be obtained when the wing incidence is greater than tail incidence (positive dihedral). For an unstable aircraft, it's the opposite - wing incidence will be less than tail incidence at equilibrium (negative dihedral). For a neutral aircraft at equilibrium, wing incidence will equal tail incidence (no dihedral).¹
Finally, answer to the question:
As stated earlier, the amount of stability that most conventional aircraft require can only be produced by moving the CG well ahead of wing AC. However, this causes the wing to produce a negative $C_m$.
To trim this out, the tail has to produce a positive $C_m$, and to do that, it needs to produce a downforce. And it needs a negative incidence to produce that downforce.
Again, this dihedral (difference in incidence) is formed as a result of making the aircraft statically stable in pitch. That answer you linked to is referring to the fact that if the aircraft is stable, it will (generally) have this LD.
Note that simply creating a large longitudinal dihedral (giving the tail a negative incidence) won't produce stability, it's the other way around. Making a stable aircraft by moving the CG forward (and then trimming it) will produce a longitudinal dihedral as a byproduct.

¹ For simplicity, this entire paragraph ignores the fact that wing modifies the flow at tail (by inducing a downwash and reducing the dynamic pressure). In practice, the only thing guranteed for a stable aircraft in equilibrium is that lift-per-unit-area at wing will always be greater than that at the tail, which generally translates to wing incidence being greater than tail incidence.
² For simplicity, this answer only considers the contributions of wing and tail. Surfaces like fuselage and engine-nacelles etc. will have their own contribution (usually destabilizing).

@sophit I stand corrected, I was wrong about longitudinal dihedral. LD is not what makes an aircraft stable, but it is the result of having a stable aircraft in equilibrium (trim). While LD can be used to analyse why $\left( \frac{dC_{m\ wing} + dC_{m\ tail}}{dα} \right) < 0$, it is not what makes that value zero. Conventional aircraft, canards and flying wings continue to have LD, but that's just the inevitable¹ byproduct of them being statically stable. LD is not what makes them stable, having CG ahead of neutral point does.  (¹ignoring downwash!) Good day!

I'm glad (I'm helped) you finally clarify this matter, better late than never  And now it's also clear why that terminology is not to be found in any technical book: it helps (maybe) people understand the concept but it is no more correct than "equal transit time"  Have a nice day you too

@sophit Kinda. They say flow acceleration causes low pressure, and incorrectly explain what causes acceleration. In reality, low pressure causes flow acceleration. I similarly stated that LD causes stability, whereas in reality, stability is what gives rise to LD. However, unlike equal transit theory, LD is a real phenomenon - the wing incidence is always greater than tail incidence. The concept isn't wrong, it's failure on my part to correctly understand this concept. Anyways, good day!

You've got the point, that's the important thing


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ADSB DF17 Altitude decoding
I develop ADSB DF17 messages decoding device.
I see that barometric height decoding differs between documents, describing the encoding standard:

https://mode-s.org/decode/content/ads-b/3-airborne-position.html
Section 5.1.

      N is 11 bit binary number with Q bit extracted.
When Q bit is 1, then formula is

25 * N - 1000 [ft]

When Q bit is 1. then formula is vaguely described, data is Gray-coded, I suppose it is:

100 * decodeGray(N) - 1000 [ft]

http://airmetar.main.jp/radio/ADS-B%20Decoding%20Guide.pdf

There is nothing about the Gray encoding, it just states, that Q bit determines which encoding factor is used (25x or 100x)
When it comes to available libraries it is getting even weirder.

pyModeS -> when Q is 1, then everything is clear, when Q is 0, results from pyModeS, 1. and 2. algorithms are completely different... I got stuck during reverse enginnering, i am not a python guy.
libmodes -> it does not even support situation when Q is 0...

What should be the proper way/algorithm?
Thanks in advance.

I never worked on ADSB, but I did on altimeters. Altimeters used binary data for large increments of altitude, such as 1000 feet. Gray code, also know as minimal change code was used for shorter increments. I n binary when going from 0111 to 1000, all four bits change, but they might not change at the same time. You might inadvertently ready 0000 in between 0111 and 1000. With gray code only 1 bit changes at a time. 0111 in gray code was 0100, 1000 in gray code was 1100. So only the MSB changed.

The gray code referenced might be Gillham Code - https://en.wikipedia.org/wiki/Gillham_code

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text:
Lifting-Line Theory - Numerical solution for a wing with dihedral angle
I'm trying to calculate lift for a wing with a dihedral angle using the Lifting-Line Theory. While using the thesis "Non-Planar Lifting-Line Theory for Fixed and Deformable Geometries" as a reference, he describes the following governing equation for a wing with dihedral:
$$\alpha(s_0) = \frac{\Gamma(s_0)}{\pi V_\infty c(s_0)}+\alpha_{L=0}(s_0)+\\ +\frac{1}{4\pi V_\infty}\int_{\frac{-b}{2}}^{\frac{b}{2}}\frac{d\Gamma / ds}{\sqrt{(y_0-y)^2+(y_0sin\phi-ysin\phi)^2}}cos(\phi-\beta)ds$$
As explained by Anderson, I know that the original LLT equation can be solved by writting the elliptic lift distribution as a Fourier series, substituting in the equation, producing matrices and solving for $A_n$:
$$\Gamma(\theta) = \Gamma_0 sin\theta = 2bV_\infty \sum_{1}^{N}A_n sin(n\theta)$$
However, I can't do the same for the equation of the dihedral, as he adds the circulation in function of the arc length of the wing. I tried manipulating the elliptic lift distribution to be in function of the arc length whilst using the thesis as reference, and arrived at:

$$z(y) = y sin\phi$$
$$s = \sqrt{y^2+z^2} = \sqrt{y^2+(y sin\phi)^2} \rightarrow y = \sqrt{\frac{s^2}{1+sin^2\phi}}$$
$$\Gamma(y) = \Gamma_0\sqrt{1-(\frac{2y}{b})^2} \rightarrow \Gamma(s) = \Gamma_0 \sqrt{1-\frac{4s^2}{(1+sin^2\phi)b^2}}$$
However, I'm unsure if it would be efficient trying to transform this equation in a Fourier series, or even if my thought process is correct. Could anyone give me some tips on how to proceed? I would be greatly appreciated!

Can we also see the thesis you refer to? There's a link to it?

@sophit Sure! Here's a link: Non-Planar Lifting-Line Theory for Fixed and Deformable Geometries.
The formulations are on chapter 3.

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Does FAA recommends specific maneuvers for complex endorsements?
I was pretty excited for a 5 hour flight in a piper arrow in a new school this coming weekend but my instructor that I have not flown with before texted me saying that he will not sign me off after one flight and would like to narrow down the time to 2 hours only. Although I was not expecting to get signed off after one flight, I canceled the flight as I didn't want to fly with an instructor with a definitive mindset of not signing off. I would have liked to fly with an instructor who'd judge me based on my performance and learning capability during the flight and ground training and then decides whether to endorse me or recommend further training. Not only I have had bad experience with one of my "time builder" instructor as a student pilot, I have heard from other pilots that some instructors can give us hard time and not be fair.
So I wanted to ask two questions.

Is there a FAA recommended list of maneuvers/training material for each endorsement? So that if a student fulfills all the requirements and fly all the maneuvers safely, he'd get endorsed?
How do we keep instructors/DPE in check? Obviously they are experienced and safer pilots but it is quite possible that in one out of hundred scenarios they are wrong and pilots receiving training are right?

It sounds like a miscommunication. It is more than likely that the instructor is following the rules for renting a complex aircraft at that school and its insurance company. If all you wanted is a complex endorsement and don’t want to rent the aircraft for solo flight, then you should make that clear to the instructor. I got my complex endorsement with an hour of flight and no ground. But, even though I have over 250 hours of complex time, I was still required to have 5 hours of instruction and 5 hours solo to be added to the insurance policy of a friends Arrow.

To answer question #1:
A complex endorsement is just that. It’s an endorsement. There are not any prescribed maneuvers to receive the endorsement. But, the instructor is holding himself liable for your performance after receiving the endorsement. Primarily, the areas where that liability may prove hazardous to the instructor, their certificate and career is on every single one of your takeoffs, landings, and go-arounds after the instructor is no longer in the aircraft. The flows and checklists for that aircraft have to be consistent and second-nature to you even without the instructor’s prompting.
A 2 hobbs-hour flight from start-up to shut-down is typically the limit of learning for most people. Two 2-hour flights along with a couple of hours of ground instruction seems reasonable for satisfying your instructors confidence in your ability. That is all conditional to your actual performance. It is also a very subjective assessment.
To answer question #2:
As far as correcting someone with more knowledge and experience than yourself, that can and should be done if (big if) they are wrong. Many things in life and aviation are subjective. There can be multiple right answers. In the case that an instructor gives you a wrong answer, be courteous and cordial in your correction. But, your answer must be backed up with fact. You should question your own answer until it can be: backed up with a regulation or recommendations out of the FAR/AIM; documented from an FAA handbook, publication, advisory circular, or other official document; referenced from the POH, AFM, owner manual, or other manufacturer source; cited from a reputable source.
If the question can not be answered objectively, the answer may only be a matter of opinion, style, or personal technique. Or, it could be a quirk of that particular aircraft or environment. Either way, it would probably be best to tailor your training to the criteria of your evaluator.
Remember, your instructor is human. They are fallible. But, so are you. And, they have been doing this aviation thing safely way longer than you. But, regulations do change. Take a slice of humble pie before you have to eat crow. You both may learn something new or improve upon something old. And remember, a good pilot is always learning.

FAR Part 61, 61.31(e) says:
(e) Additional training required for operating complex airplanes.
(1) Except as provided in paragraph (e)(2) of this section, no person may act as pilot in
command of a complex airplane, unless the person has—
(i) Received and logged ground and flight training from an authorized instructor in a complex airplane, or in a full flight simulator or flight training device that is representative of a complex airplane, and has been found proficient in the operation and systems of the airplane; and
(ii) Received a one-time endorsement in the pilot’s logbook from an authorized instructor who certifies the person is proficient to operate a complex airplane.
(2) The training and endorsement required by paragraph (e)(1) of this section is not required if—
(i) The person has logged flight time as pilot in command of a complex airplane, or in a full flight simulator or flight training device that is representative of a complex airplane prior to August 4, 1997; or
(ii) The person has received ground and flight training under an approved
training program and has satisfactorily completed a competency check under
§135.293 of this chapter in a complex airplane, or in a full flight simulator or
flight training device that is representative of a complex airplane which must
be documented in the pilot’s logbook or training record.
I cannot find anything that provides a list of maneuvers you must demonstrate.
Complex is defined as plane with constant speed propellor, flaps, retractable landing gear.
Lots of planes have 2 out 3 of those, so you are really just adding the retractable landing gear to your repertoire. Do you need a 5 hour flight to demo enough take off and landings to perform that at the appropriate speeds and for the school to be confident you not belly it in and cost them an engine tear down, prop replacement, belly damage repair, etc?  Or for you to know enough emergency procedures to get the gear extended if something fails and it wont come down with normal procedures.
Probably not, hence the 2 hour checkout request from the instructor.

May be I wasn't clear @CrossRoads, the 2 hour checkout from instructor was just got get familiar with me and not endorse me. He specifically said he won't be endorsing me unless he went out few times with me.

Seems reasonable to me.  He works for the flight school, yes? He's protecting your life, his livelihood, and their airplane and insurance premiums (and loss of airplane rental time while it is being repaired, assuming it can be).  The saying goes there are 2 types of pilots who fly retractables - those who have and those who will (land gear up).  You don't want to be in the first category too quickly.

How is he protecting? I wasn't going to get checked out in their airplanes. Complex airplanes at my local school only 2 minutes away has hour limit so I can't fly. This school is 60 miles away. Just wanted to go there for endorsement. Which could be done in one day, two flights, three days. But the mindset was questionable here.

@sahmad -  If all you want to do is to just get some time in a complex airplane, do it. If you want to get an endorsement without an expectation of flying solo, call around to other local instructors. You may find that it is difficult to find a complex, single-engine trainer available. If you are looking for a complex airplane to fly solo, expect a 5 hour checkout. My local flying club requires a minimum of 5 hours with the club instructor on each complex airplane regardless if you already have a complex endorsement or have been checked out in another club aircraft.


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How did the MiG-25M prevent a flameout?
How do afterburning turbofan or turbojet planes prevent or minimize the risk of a flameout at nearspace altitudes such as Fedotov's Mig-25M that went higher than 115K ft (35 km) two times? At these altitudes the MiG's engines were shut down already (the plane was floating on inertia like a suborbital spacecraft), but they must have been shut down quite high anyway, 80 or 90 thousand feet perhaps. Is there a way we can get to know accurately the highest possible altitude for a particular jet engine to run? Is there a way for the pilot to know when they could shut the engine down before it does on its own in a flameout?

you seem to have created multiple accounts. Be aware that if you lost access to an account, you can ask for it to be merged to your current one.

@Federico Thank you but I don't really care. I'm unregistered. A new account forms when posting anything on another SE or when deleting the browser's cookies.

I am just mentioning in case you are not aware. registering allows you to keep the reputation and access to old posts you have made, in turn giving you access to more tools. You are more than welcome to ignore that if you wish.

The engine flamed out on the way up. Inertia alone carried the Ye-155M to its record altitude. Its high speed and its high supersonic load factor of 5 allowed to fly a zoom climb where the pilot pulls into a vertical climb at an altitude which still supports a pull-up maneuver (which requires a flight speed near the airplane's top speed of Mach 3 for best results). This page lists the maximum altitude for sustained flight of the Ye-155M at 24,200 m.
The engine, a R15BF2-300, started life in 1958 as the power plant of several unmanned, supersonic reconnaissance airplanes (R15K, used on Tu-151 and -153 Yastreb). For this purpose it was of rather simple design with only 5 compressor stages and no bleed air for turbine cooling. Later it was  upgraded for manned airplanes like the Ye-150 family, being the only low-cost solution to a high-altitude supersonic reconnaissance airplane. For the record-setting Ye-155M a sixth compressor stage was added, turbine temperature increased and reliability improved.
You might ask why so few compressor stages were sufficient for high altitude operation. The reason is supersonic flight: Most compression is from ram pressure in the intake, so the engine does not have to add much compression on top - or if it would, compressor exit temperature would be too high to add much energy from combustion before turbine entry temperature is exceeded.

Years ago read Igor Belenko's story about his defection to Japan with a Mig 25, "Mig Pilot".  He ended up flying for one of the US majors, American Airlines I think.

@JohnK you surely mean Viktor Belenko, right?

Yes Igor was some other guy lol.


text:
Sea-Going Ground-Effect Vehicle
Is a Ground-effect vehicle (a.k.a. ekranoplan) practical in seas near the coast (e.g. the US East Coast, West Coast, and Gulf of Mexico)?
I'm looking for a practicality answer, like, nope, cannot be done, yes, it can, or, we don't really know yet.

Problem is you have all these vehicles flying around at high speed at the same altitude.  If it got busy enough you'd eventually get them running into each other.

This is being done right now (article from April 2022): "A prototype sea-to-air passenger craft has clocked more than 35 knots (70kph) in sea trials off Tampa, Florida this month - and a New Zealand company is today announcing the $700 million purchase of some of the first commercial production models." https://www.newsroom.co.nz/nz-investors-order-25-foiling-electric-gliders-for-high-speed-coastal-commuting

can you define what you mean by "practical"?

As far as I know no ground-effect vehicle has ever operated regularly on the open ocean.

@Federico: Sure, practical = can capture most of the benefits of travel in a lake environment or Caspian Sea (like ekratoplan), but in an ocean environment near the coast.

70kph...is slower than a very fast cyclist.


text:
Is a supersonic ekranoplan possible? Would mach 1+ would disable ground effect?
An ekranoplan takes advantage of ground effect to make it "fly" very economically.
Is it possible for an eraknoplan to fly at the speed of sound, how does ground effect work at such high speeds?
How a sonic boom works
I'm thinking theoretically; assuming that the wind won't push it in any way, and that there is a huge open expanse for it to go through.

A car has broken the speed of sound, so it's certainly possible. Hitting a wave at supersonic speeds would be pretty catastrophic.

Not to mention how few places there are with long enough stretches of calm waters to justify flying that fast.  Monster lakes like Superior or Baikal tend to be almost as rough as large bays.  Rivers aren't straight enough for transsonic speeds.  And the ocean, of course, can have waves several meters high even in locally calm conditions, due to distant storms.

The problem are reflected shockwaves which will upset longitudinal trim.

@PeterKämpf the frequency of the reflected shockwaves vs the damping of longitudinal trim would make a great study (with scale models first).

@PeterKämpf your comment is begging to be expounded into an answer

No, supersonic flow will still show lower drag in ground effect. However, wave drag will become so high that the performance advantage of flying in ground effect would become small. Just fly a little slower and performance will improve more than can be achieved with flying lower. Also, pressure loads on the structure will go up as height is reduced.
From "Transonic and supersonic ground effect aerodynamics" by G. Doig, published in Progress in Aerospace Sciences 69, August 2014, pages 1-28:

it can be seen that approaching Mach 1 will severely reduce L/D. Unfortunately, this measurement does not extend into pure supersonic flow, but the trend should already be obvious. However, flying close to the ground still gives a performance advantage at equal Mach numbers.
NASA-TM-X-611 has now been declassified and shows pressure profiles along the length of both an F-104 and an F-105D in ground effect. From the linked paper:

The tests were primarily concerned with shockwave potential to damage structures and break windows, etc., but provide almost certainly the first comprehensive measurements of strong shock/ground interaction from a low-altitude supersonic overflight.

Even at moderate height, the peak pressures start to go up as height is reduced. This means that supersonic flight in ground effect requires a strong structure on the lower side! Altitudes were still outside of those which cause shocks to be reflected back to the airplane, and the test pilots did not report any control problems. However, when flying very close to the ground, shocks will be reflected by it and hit the lower side of the airplane, causing changes in pitch:

Studies of NATO 5.56 mm projectiles at Mach 2.4 in ground effect show this.  Since the projectiles were spinning, the changed aerodynamic forces only produced a lateral precession. Interestingly, when the reflected bow shock is reflected into the near wake at heights between 1 and 1.3 times diameter, it lowered drag by raising back pressure.

The projectile was discovered to experience a marked pitching moment, initially nose-downwards at the high end of type C (h/d = 0.75) reflection interactions, then more strongly nose-up with continued decreasing ground clearance, as the pressure distribution over the rear of the projectile was increasiingly distorted by the build up of high pressure behind the series of shock/ground interactions.

Wake and vortex ring structure behind a Mach 2.4 projectile for various ground clearances. All figures and citations copied from the cited article. For the original sources, please refer to the references section in that paper.

No, supersonic speed would not disable ground effect.
Ground effect is caused by air that is deflected downwards by the wing reflecting its energy back up to the bottom of the wing.

evidence that ground effect interacts with downwash is shown by the fact that ground effect lowers stall AoA.

Higher speeds only increase the energy of air being deflected downwards, and at supersonic speeds, a well defined shockwave also develops.  Designers of the XB-70 took advantage of this by adding a highly Ekranoplane-like anhedral to the movable delta wingtips in order to "ride" the higher air pressure underneath at supersonic speeds, generating additional lift.
There would be considerable safety concerns attempting to develop this concept, as unknown obstacles would be almost impossible to avoid at supersonic speeds.  Low level sonic boom would also be a concern.


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Why increase chord length for supersonic aircraft (swept vs delta wings)?
TL/DR
For supersonic transport and a given sweep angle, why are delta wings used instead of swept wings? If you write the drag equations, a sweep of $55^{\circ}$ seems sufficient with no modification to chord length. Why then do almost all supersonic passenger aircraft have delta wings?
Calculation
Let's start w/ the weight of a Boeing 737, then hypothetically set the cruise altitude to ~24000 m, where air density is 10% of 12000 m (the 737's normal cruise altitude). I'll use MTOW here as the weight of the aircraft at cruise (which is high, but will provide an upper bound on the calculations). To generate enough lift, we have:
$$ F_L = \frac{\rho v^2 s C_L}{2}$$ where $s$ = surface area. If we're a high altitude where $\rho$ is 10% of what it normally is, then to generate the same lift $F_L$, you need to either change $v, C_L, $ or $s$. I wrote a Python script to iterate through the following:

Increasing surface area from $s$ to $2s$
Increasing lift coefficient $C_L$ from $0.5$ to $1.2$

The resulting graph looks like this:

$$ $$
The x-axis is $C_L$. From this graph, we can calculate new values for $s$ and $C_L$, and the resulting $F_L$. $F_L$ is much lower than the lift $F_{L737}$ required to lift a 737 , hence we will need to increase $v$ significantly. Using our new values for $s$ and $C_L$, the velocity required to generate enough lift is
$$ v = \sqrt{\frac{2F_{L737}}{\rho s C_L}} $$
The velocity required is about Mach 1.5. Now we calculate the drag force, which will determine the force required to keep this plane in the air.
$$ F_D = \frac{\rho v^2 s}{2} \left( c_{Di} + c_{Dp} + c_{Dw} \right) $$
where $c_{Di}, c_{Dp}$, and $c_{Dw}$ are the coefficients of induced, parasitic, and wave drag, respectively.
$$ c_{Di} = \frac{C_L^2}{2 \pi A_R e} $$
$$ c_{Dp} = constant $$
$$ c_{Dw} = 20(M - M_{cr})^4 $$
$A_R$ is the aspect ratio, $M$ is simply $v$ expressed as Mach, and $M_{cr}$ is the critical Mach number. This is where I don't get delta wings vs swept wings.
$$ $$
Using the Korn equation (p.18 here), $c_{Dw}$ is:
$$c_{Dw} = 20\left( M +\sqrt[3]{\frac{0.1}{80}}+\frac{t/l}{\cos^2\lambda}+\frac{c_i}{10 \cos^3 \lambda} - \frac{k}{\cos\lambda} \right)^4 $$
$l$ is the chord length and $\lambda$ is sweep angle. $c_{Di}$, when $A_R = \frac{b}{l}$ is:
$$ c_{Di} = \frac{C_L^2l}{2 \pi b e} $$
It seems that just sweeping the wings back (increasing $\lambda$ to $55^{\circ}$) would be sufficient to reduce wave drag without reducing $l$. Why then do supersonic aircraft have delta wings where $l$ is quite large? It seems like chord length $l$ does not matter too much in supersonic, but the resulting $A_R$ changes would cause pretty inefficient subsonic flight. Is there a component of drag here for the wing that I'm missing?
Open to all answers, but mathematically justified would be awesome.

Have you considered the structural implications of a wider vs narrower wingspan of the same area?

Not specifically yet, though current numbers are based on t/c ratio of Boeing 737. From literature, low aspect ratio wings like those used on fighter jets can have t/c as low as 0.05.

Wild guess.  It may have more to do with mounting the engines underneath, taking advantage of all that higher pressure air there.  Also, keep in mind a passenger supersonic hasn't been developed in 60 years.  Deltas are stable, conservative, and reliable.  Perhaps, the next one may look like a B-1.

The Korn equation is for estimating the Mach number where the transsonic drag rise starts. It cannot be used for supersonic flow. Besides, deltas allow to pack much more fuel at low wave drag which is a winning combination for supersonic flight.

Does this answer your question?

@PeterKämpf that was extremely helpful. I was extending Korn to model supersonic drag, which is completely incorrect. Am currently looking into modeling supersonic drag, but will save this for another question once I've taken a crack at it.

Supersonic wings needs to be thin, which makes it weaker, and long chord counteracts that. 2) Larger wing area increases lift.

Open to all answers, but mathematically justified would be awesome.

Aerodynamics is what keeps aeroplanes in the air, and is ruled by physics. The equations used in OP are simplified physics equations, and the main thing to observe is if the simplified equation is applicable to the case under consideration.

The first equation used in OP, usually written as $F_L = C_L \cdot ½ \rho V^2 \cdot S$, is valid in low sub-sonic conditions, when the compressibility of air can be neglected - for airspeeds lower than Mach 0.4 - 0.5. In these conditions $C_L$ is reasonably independent of dynamic pressure, at higher airspeeds this is not the case anymore and $C_L$ will change with M. As can be seen in the graph underneath, from TU Delft course Introduction Supersonic Aerodynamics D 25-A by Prof. dr. -Ing Erdmann. The equation is of very limited value in transsonic and supersonic cases where compressibility of air plays a role, and should not be used without considering the validity conditions.

The Korn equation is a generalised equation that can be used for design calculation of expected drag of the wing. In transsonic circumstances, as pointed out by @PeterKämpf in a comment, so with M between 0.8 and 1.2. Not to be used at M > 1.2, when the shockwaves are the main determinant of supersonic drag.

Pic earlier used in this answer
What is the use of delta wings for supersonic aeroplanes? It keeps the wing inside the shock cone - would it be sticking out of the shock cone, additional compressibility drag would be created at wing bit that sticks out.

Why delta instead of swept? Image above from Torenbeek, first used in this answer, shows four bombers with planform shapes ranging from pure sweep to pure delta. Consider a strongly swept wing @ 55°: there is a triangular gap between the aft wing edge and the fuselage. Fill this space up with wing structure to obtain a delta wing, and:

Wing area increases, reducing wing loading.
Structural stiffness increases, reducing flutter & wing twist problems.

Note that the depicted aeroplanes are high subsonic ones - the delta wing was considered advantageous in this cruise speed region!
Pic source
One supersonic fighter jet with straight swept wings instead of delta wings, The English Electric Lightning, looks like a delta with the inner aft triangle of the wing shifted further aft, to become the swept horizontal tail.

As sweep reduces the effective velocity over the wing, it reduces drag. However, the same is true for lift, see the image below:

Source
With increasing sweep the effective airfoil becomes thinner, and lift reduces.
As a consequence of this, the surface area needs to increase to make landing performance acceptable.
You can do this by increasing the chord of a swept wing, or more structurally efficient, by creating a delta wing.

Meaning that you can either have a really long swept wing (impractical structurally) or a delta. This makes a good amount of sense. It also seems that delta wings have superior supersonic performance, but I need to take a crack at the math here. I think it has something to do with reducing wave drag - allowing for lower angle of attack, but will be working on the math. Will probably save this for another question. Thanks!

Using a delta wing instead of a highly swept wing for supersonic aircraft has positive effects both from an aerodynamic and a structural point of view.

From an aerodynamic point of view, the biggest source of drag in supersonic flight is the “wave drag” i.e. the drag generated by the shocks (waves). It can be derived (mathematically from the small-disturbance potential equation) that it exists an optimal distribution of the cross-sectional area of the aircraft which minimises the wave drag. The cross-sectional area is defined as the area obtained slicing the aircraft from nose to tail, as depicted in this picture:

Definition of cross-sectional area. Source: https://upload.wikimedia.org/wikipedia/commons/thumb/6/6f/Area_rule_unifilar_drawing.svg/1092px-Area_rule_unifilar_drawing.svg.png
This ideal distribution is called "Whitcomb area rule": moving from nose to tail, the cross-sectional area smoothly increases from zero at the nose till a maximum at the center and afterward gently decreases toward zero at the tail. How the cross-sectional area is distributed among fuselage, wing and other components doesn't matter. This picture shows how the area rule was used on the fuselage of the F-102 to regularise its cross-section in order it to be as smooth as possible:

How the area rule works in practice. Source: http://www.aerospaceweb.org/question/aerodynamics/q0104.shtml
So why a delta wing instead of a highly swept wing? A delta wing (eventually combined with a small reduction of the fuselage’s cross-section toward the center, like in the F-102 of the previous picture or in the most recent iteration of Boom Overture) helps in achieving this smooth change of the cross-section. Note that this result could possibly be achieved also with a swept wing if the swept is forward or adding so called anti-shock bodies.
A second important aerodynamic aspect is to have the entire wing inside the Mach cone. When the wing lies inside the Mach cone, the component of the speed perpendicular to the leading edge is always subsonic and that not only avoids the generation of wave drag but mitigates the Mach tuck as well. The following slide depicts this aspect:

The right wing lies completely inside the Mach cone giving less drag at supersonic speeds. In the slides also some other nice examples are presented. Source:
https://www.slideserve.com/grace/mae-3241-aerodynamics-and-flight-mechanics
For an aircraft flying at Mach 2 (like the Concorde), the line connecting the nose with the far end of the wing should have a sweep bigger than some 60° since the Mach cone has an angle $\mu$ of 30°.
So again, why a delta wing instead of a highly swept wing? The lifting area of a delta wing is concentrated toward the fuselage and therefore the wingspan can be reduced in respect of a highly swept wing, the area being the same. This gives a more compact wing which is more easily fit inside the boundary of the Mach cone.

From a structural point of view the delta wing is more efficient than a simply swept wing. This cutaway of a F-111 helps in visualising this:

F-111 cutaway. Source: https://www.pinterest.de/pin/391672498829049341/
The F-111 was a supersonic aircraft with variable swept wing. On the cutaway I highlighted in red the load-path, from wingtip till fuselage, when the wing was swept backward of some 70°. What would have been the same load-path if the wing were a delta instead? The much shorter blue line gives the answer.
So again, why a delta wing instead of a highly swept wing? Because the spars can leave the fuselage perpendicularly to it giving a much shorter and more structurally efficient (i.e. lighter) “tip-to-tip” load path. Just as an another example, this cutaway shows how the spars of the Concorde looked like:

Concorde cutaway. Source: https://www.flightglobalimages.com/cutaways/civil-aviation-1949-present-cutaways/bae-concorde-cutaway-drawing-1571101.html?prodid=693

Obviously a delta wing has also an impact on other aspects: in general it provides more room for fuel, engines, avionics and payload.
Furthermore the weight of the wing is closer to the roll-axis, giving a more agile (combat) aircraft.
As usual in the aerospace world, it is always a matter of compromises :-)

Not clear how this answers the stated question, which is NOT "why delta" but "why delta instead of swept"? In what ways does a swept wing not gain the same benefits as a delta?

@RalphJ: the question asks "why increase chord length" and "Why then do almost all supersonic passenger aircraft have delta wings?". The answer is because it gives a better cross-sectional area distribution + it keeps the wing inside the Mach cone + it gives better structural path + it gives more space for fuel. All things listed in my answer. Obviously all the "better" things listed are in respect to a swept wing as asked in the question. Anyway, I'm going to update my answer with your comment


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Why do autopilot controllers use $\theta$ as the input signal rather than $\gamma$?
In particular, referencing how the Matlab/Simulink controller here uses $\theta$ (absolute pitch angle), rather than $\gamma$ (flight path angle). (Autopilot controller starts at 59:26)

It seems if you wanted to control flight path angle, your input signal should be $\gamma$. I'm sure I am missing something here.

If you are talking about angle of climb/descent, then according to Principles of Flight by Oxford (this page), it is appropriate to use both $θ$ and $γ$ to denote that...

@Aditya Sharma The symbols of the pitch angle $\theta$ and the flight path angle $\gamma$ are defined and accepted throughout the community. For example the DIN9300 (unfortunately this norm is not open source) or in pretty much every book, e.g. Flight Dznamics principles by Michael Cook.

@MichaelHall, I agree. I would propose a corresponding edit of the question, but I only seem to be able to directly edit it (without approval)

@MichaelHall I proposed the edit to the community

Just fixed. Thanks!

You are right, your input should be $\gamma$. However the control of $\theta$ is important. To combine both variables, typically a nested control structure is adopted.
For this, two control loops, an outer loop and and inner loop is connected in series. I have drawn a short sketch

Here you can see that the climb angle error $\Delta \gamma = \gamma_{cmd} - \gamma$ (the difference between the measured and commanded climb angle) is fed to the outer loop, which calculates a pitch command. The pitch error is then calculated by $\Delta \theta = \theta_{cmd} - \theta$. This pitch command is then fed to the inner loop controller which finally computes an elevator deflection for the aircraft.
In most applications the outer and inner loops are simple PID controllers, which are gain-scheduled over the flight envelope. Additionally it should be noted that the inner loops itself is often divided into a rate controller (controlling the rotational rate $q$ for the pitch axis) and and attitude controller controlling $\theta$.
However, this should serve as a general overview how controlling pitch and the flight path angle at the same time can be achieved.

how does this differ at all from my answer? you're simply limiting the choice of external control loop to a $\gamma$ one, where that limitation is not needed at all.

It includes a nice picture.

@MichaelHall then keep your pictures

@Federico, chill out dude… competing answers is how this site works.  You ought to know that, aren’t you a moderator?


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What is the minimum runway required for an a320 under 72,000 kg
How much is the absolute minimum runway required for takeoff and landing? How much could you reduce the takeoff/landing distance using various methods (eg-standing takeoff)? I'm talking about an A320 under 72,000 kg.
Note: This is at sea level and at normal temperature.
Many sources will provide you with the minimum runway requirements of an aircraft

(eg Wikipedia: Takeoff:2,100 m (6,900 ft) Landing:1,500 m (4,900 ft)

I want to ask how much could you reduce the minimum requirements further by using non-standard methods (such as holding the brakes until the engine spools up, taking off into headwind)

If you’re going down the headwind path then you need exactly zero feet of runway.

Holding the brakes is more “standard” than you’d think. The most ideal scenario would be a direct headwind on a dry runway on a dry, cold day with high atm pressure (lower pressure alt). I don’t know if the test pilots had all these conditions, but the results of the max performance landings would be a good reference (or the shortest landing distance ever noted). A thorough google search would be needed, at the least. On top everything there’s pilot’s skill level. If you want to know theoretical values, feel free to dig in to the A320 POH.

By normal temperature do you meant standard day?  (29.92 inches mercury, sea level, 15° C?

https://youtu.be/cHhZwvdRR5c  And https://youtu.be/602wpcM6EvU

LATAM operate A320 in Rio(SBRJ)... 1320 meters and the sugar lof ahead.

The parameter which will influence take-off distance the most is weight.
The figures you quoted are take-off distances at MTOW. But if you reduce the take-off weight, you can use significantly shorter runways.
https://www.airbus.com/sites/g/files/jlcbta136/files/2021-11/Airbus-Commercial-Aircraft-AC-A320.pdf section 3-3-2 gives indications. Here's an example for an A320-200 with CFM56 engines:

As you can see, reducing the take-off weight as much as you can could get you down to about 1200m for take-off. But at sea-level, that means a take-off weight around 64t, 14t below MTOW.
At 72t (which is really quite close to MTOW), you'd probably be around 1600-1700m.
For reference, Gibraltar airport has only a 1776m runway, and still sees A320s flying in and out every day, and I'm pretty sure Easyjet doesn't fly theirs half-empty! That means they have to limit the amount of fuel, so probably no flying to Sharm El Sheikh from there, but the UK seems to be in range.
That's why BA's LCY-JFK flight using A318s stopped at Shannon (only in that direction): the short 1500m runway does not allow to take enough fuel for the whole transatlantic flight (the return flight JFK-LCY was non-stop as there was no limit on fuel taking-off from JFK -- though this was also partially helped by the jet stream). Remember that those A318s were all-business-class so also had a significantly lower payload than usual, which shows the importance of fuel.
Reducing take-off weight is probably the only reliable method of reducing take-off distance, so you can use it for scheduled operations. Other parameters are either intrinsic to the airport (nature of the runway, altitude...) or you have no control over it (wind speed and direction, air pressure...).
If the question is not about regular scheduled operations but rather a one-off event (e.g. "had to land on a very short runway in an emergency, how do I get out of there now?"), then you could indeed wait for some of the better conditions, but I'm pretty sure the first thing you'll do is reduce weight first. No payload, and just the amount of fuel required to get to the closest "normal" airfield. Only in the case that wouldn't be enough would one need to be more creative.


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How to arrive and depart surface class E airspace
I fly VFR. I found all the information about the airspaces and I believe I understood everything. What I'm still struggling to find is how to properly arrive and depart surface class E airspace surrounding certain airports. I never practiced that before and non of my instructors ever mentioned how to do that. What I understood from class E descriptions, I have to get authorization to enter surface class E airspace. How to do that? What frequency to use? Where to find it? What exactly should I say? What to expect in reply? Can I do touch an goes there? Is it possible that my request will be declined? Does anybody have a youtube URL showing an example how to do that?

You are not required to get authorization to enter class E airspace.  This is fundamental ground school and pre-solo 101 information you should be discussing with your CFI.

Welcome to aviation.SE! Are you asking about the US, or another country?

It would also be helpful if you can give us a specific example of an airport that you're asking about.

Sorry I didn't mentioned that - yes, I'm asking about the US. Accordingly to this https://www.faa.gov/uas/resources/events_calendar/archive/2019_uas_symposium/media/How_To_Understand_and_Operate_in_Class_E.pdf, see surface class E Type E2 Authorization Needed. I found other sources stating the same. KTPL for example. How to properly get that authorization?

The document in that link is for Part 107 operations. If you are flying a manned aircraft VFR, you are not flying Part 107. You are flying Part 91 at the very least or possibly Part 103.

I didn't realize that was only for "unmanned" aircraft. Thank you for explanation.

Class E airspace has no requirements for communication when flying VFR. You only have to maintain two-way communication in Class E airspace when flying IFR. All airports where there is an operating control tower requires two-way communication within 4 Nautical Miles of the airport housing the control tower at and below 2500 feet AGL at a minimum, regardless of airspace. You basically treat all airports with an operating control tower like a Class D airport. A, B and C airspace rules are even more strict. You maintain that communication either through radio or light gun signals.

§91.127 Operating on or in the vicinity of an airport in Class E airspace.
(a) Unless otherwise required by part 93 of this chapter or unless otherwise authorized or required by the ATC facility having jurisdiction over the Class E airspace area, each person operating an aircraft on or in the vicinity of an airport in a Class E airspace area must comply with the requirements of §91.126.
(b) Departures. Each pilot of an aircraft must comply with any traffic patterns established for that airport in part 93 of this chapter.
(c) Communications with control towers. Unless otherwise authorized or required by ATC, no person may operate an aircraft to, from, through, or on an airport having an operational control tower unless two-way radio communications are maintained between that aircraft and the control tower. Communications must be established prior to 4 nautical miles from the airport, up to and including 2,500 feet AGL. However, if the aircraft radio fails in flight, the pilot in command may operate that aircraft and land if weather conditions are at or above basic VFR weather minimums, visual contact with the tower is maintained, and a clearance to land is received. If the aircraft radio fails while in flight under IFR, the pilot must comply with §91.185.

§91.126 Operating on or in the vicinity of an airport in Class G airspace.
(a) General. Unless otherwise authorized or required, each person operating an aircraft on or in the vicinity of an airport in a Class G airspace area must comply with the requirements of this section.
(b) Direction of turns. When approaching to land at an airport without an operating control tower in Class G airspace—
(1) Each pilot of an airplane must make all turns of that airplane to the left unless the airport displays approved light signals or visual markings indicating that turns should be made to the right, in which case the pilot must make all turns to the right; and
(2) Each pilot of a helicopter or a powered parachute must avoid the flow of fixed-wing aircraft.
(c) Flap settings. Except when necessary for training or certification, the pilot in command of a civil turbojet-powered aircraft must use, as a final flap setting, the minimum certificated landing flap setting set forth in the approved performance information in the Airplane Flight Manual for the applicable conditions. However, each pilot in command has the final authority and responsibility for the safe operation of the pilot’s airplane, and may use a different flap setting for that airplane if the pilot determines that it is necessary in the interest of safety.
(d) Communications with control towers. Unless otherwise authorized or required by ATC, no person may operate an aircraft to, from, through, or on an airport having an operational control tower unless two-way radio communications are maintained between that aircraft and the control tower. Communications must be established prior to 4 nautical miles from the airport, up to and including 2,500 feet AGL. However, if the aircraft radio fails in flight, the pilot in command may operate that aircraft and land if weather conditions are at or above basic VFR weather minimums, visual contact with the tower is maintained, and a clearance to land is received. If the aircraft radio fails while in flight under IFR, the pilot must comply with §91.185.


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Need communication example departing and arriving uncontrolled airport within class D
I'm looking for a communication pattern to depart and arrive an uncontrolled airport within class D airspace VFR. One example can be F69 which is within KADS. This case looks even trickier because this airspace is right under class B. It seems to me it is possible to go under class B especially if going east. Would it be recommended considering how busy class B can be or it is actually a viable option to go from F69 into class D and then class B? It would be great if somebody could post links to videos. I'd like to know exactly what to tell and what to expect in reply for both cases departing and arriving.

I fly out of Addison, McKinney, Mesquite, Denton, Rockwall, Sulphur Springs and Aero Country a lot. I am also friends with the A&P located at Airpark. If you would like a safety pilot who knows the local area, maybe we can hook up.
A few things to remember when operating out of airpark:

The runway is barely maintained. But, it is not perfectly smooth. Treat it like a soft field.
When taxiing anywhere besides the runway, you will be taxiing on turf. Protect your nose wheel.
In the daytime watch out for chickens.
At night, beware of coyotes.
Do not park, walk, or stand under the concrete bikini hangars. They are off-limits due to instability and collapse.
The windsock is in disrepair. Do not trust it. Look for other nearby flags and smoke from local bbq restaurants.
The runway has non-standard markings
The runway is a lot shorter than you think due to displaced thresholds at both ends due to buildings, fences, and trees.
Make all of your radio calls on the CTAF while taxiing. Make a few CTAF radio calls while in the pattern and right before taking the runway for takeoff. All of your other radio calls will be on Addison or Fort Worth frequencies. There is very little traffic at Airpark (1-2 aircraft a day). All of the other traffic in the area will be on Addison or Fort Worth frequencies.
The runway is aligned with the runway at Addison. You will be on a very similar final and departure leg.
Addison is only 3 miles away. You can almost see the Airpark area from the Addison Tower.
Because of Addison Tower’s Proximity to Airpark, contacting them as soon as practicable means
as soon as you start the engine.
Addison Tower does not like hearing from pilots directly as an initial radio call. They prefer pilots to contact Ground first and get transferred to them via a frequency change. At Airpark, it would be best to get Clearance Delivery and/or flight following first through Fort Worth Approach. Then, contact Addison Tower.
Inbound to the Addison Class D, you are required to listen to the ATIS first. There will be a frequency given over ATIS to contact Fort Worth Approach. You must contact Fort Worth Approach before contacting Addison Tower. Period! Trust me when I say Addison Tower will not like it if they are your first point of contact.
Addison is an extremely busy Class D. It caters to jet and turboprop business aviation. If it did not have DFW and Love Field about 10 miles away, it would probably be a Class C or at least a TRSA.
There can be as much as a 15 minute and sometimes up to an hour wait in the run-up area for clearance to takeoff. And, they will sometimes utilize a line up and wait clearance. This can affect your takeoff clearance at Airpark.
Good visual reference points for Airpark are the Dallas North Tollway and the President George Bush Tollway. Airpark is located in their Northwest corner.
The Airpark area is extremely congested. Climb out to a safe altitude as soon as possible. If you have an engine out upon takeoff you have very few options. None of them good. To the North, you have a green space on the East side of the Dallas North Tollway. To the South, you have Addison Airport.
Do not contact the Airpark airport owner, operator, or manager listed as points of contact. They are anti-aviation real estate developers who bought the land as an investment. They will try to steer you clear of Airpark as much as possible. Their acquisition of Airpark is very similar to Meigs Field in Chicago. They bulldozed the operating FBO and fueling area. But, it took a court order to keep them from bulldozing the runway.
Most pilots choose to go North and East to practice maneuvers, to practice instrument approaches, and to beat up the traffic pattern at other airports.
If you have at least a private pilot certificate, do not worry or be intimidated by the Class B. If you are flying to, from, through, or above the Addison airspace, you will need to be in two-way communication with ATC anyway. Just ask for clearance through the Bravo.

Thank you for this detailed answer, I'm impressed with the number on nuances about F69. I'm lucky I got an answer from somebody that familiar with the area. All that makes me to rethink my initial intensions. Event though F69 is the closes to my ground target, I might better consider other destination like T31 further from all complexities.

@AlexK -  In which cardinal direction is your ground target? There are better choices for transient traffic than Aero Country. No offense to Aero Country. They are very well maintained. But both them and Airpark are residential, live-in airfield communities. You may be better off at a full service facility like McKinney (KTKI). The city owned and run FBO has crew cars, rental cars, a pilot lounge with a theatre, shower, full kitchen (upstairs), flight planning room, and free ice cream in the summer. It was voted #1 FBO in the country. Plus, it’s close to downtown McKinney. And, there’s KADS.

After your description of the traffic at KADS I simply assumed that the situation at KTKI will be similar and named another non-towered airport at the area but if you recommend KTKI I believe it will work, it is not much further. The actual ground target is just about 3 may be 4 miles west of KADS, so location-wise it is the best but my reasoning is this, instead of waiting for 15 or 60 minutes for clearance to take off I better use a smaller airport further away and spend less time than that on the ground to get where I need to.

@AlexK - TKI does not have nearly as much traffic as ADS. ADS’s traffic can be summed up in the name of its leading FBO, Millionaire. TKI is much more laid back, catering to student as well as corporate traffic. Plus, it is near the start of a major SW tollway. There is no need to contact approach inbound to TKI unless you are IFR. It sits comfortably under the very edge of the Bravo. There are plenty of open fields N & E for emergencies. And, T31 to its W. You can get nearly immediate takeoff clearances at TKI. At its busiest, you may wait 15 minutes. Visit Hutchins BBQ while you are there.

Another option is Mesquite (KHQZ). They are close to IH635. It is a major East-West thoroughfare. Denton is near  IH35E, the main North-South freeway. They have good crew cars as well.

If you are departing an uncontrolled airport within the surface area of a controlled airport, then you are required to contact the latter airport's control tower as soon as practical.
In the particular case of F69 and ADS, since the former is almost directly under the instrument approach path to the latter, you will also want to turn east (i.e. away from the approach) as soon as practical when the winds are from the south, which is the prevailing direction.
Note that this doesn't remove your obligation to make CTAF calls at F69, and be very wary of inbound aircraft not switching from ADS tower to F69 CTAF until short final. Transmit early and often, and once you enter the runway, do not dilly-dally there.
Once you're off the ground, inform ADS tower "Addison Tower, N12345, off Airpark, VFR eastbound." They will probably give you a local squawk code in return. Depending on how busy they are, they may be able to set up Flight Following, or they may tell you to request it from Departure.
When returning, ADS requires you to contact Approach first for sequencing. They will switch you to Addison tower; include "landing Airpark" in your check-in call as a reminder, and they will switch you to CTAF on final. Beware of aircraft already on the runway (because they didn't know you were coming) and be prepared to go around if needed—and as soon as practical, switch back to Addison tower for resequencing.
Working both frequencies at the same time may seem safer, but don't attempt it single-pilot unless you're absolutely sure it won't distract you during a critical phase of flight. It's not worth potentially forgetting to put your gear down or under/overrunning the runway because you came in too slow/fast.

There should be no reason or need to enter the bravo.
I would listen to KADS ATIS to determine winds and direction of landing, then simply give tower a call and tell them you are inbound to F69.
For a specific example:  “Addison tower, Skyhawk N123Y 10 miles northeast, information Bravo, inbound for Air Park (F69).”  They will probably come back and ask you to report entering 45 for the downwind leg or something.
For additional preflight planning assurance, give tower a call on the land line and ask about any local procedures for F69.  They are sure to give you a better answer than I can!

Generally, any entry into the Addison Class D would be by radar vectors. You have to contact Fort Worth approach prior to contacting Addison Tower. The ATIS will direct you to do so. Approach usually gets you lined up with final before handing you off to Tower with a freq change. During slow times, approach may hand you off early. If so, Tower May let you go directly into the Airpark traffic pattern. But, it is very rare to find a slow period in daytime VFR in Addison. Night landings at Airpark are sketchy at best. They are very similar to night soft field landings.

There are some really excellent answers and comments here from people with first hand knowledge of this specific airport.  Very helpful to the OP.  But, the question was a general one, using this airport as an example, and generally speaking radar vectors aren't needed to get to a secondary airfield in Class D under VFR.

Granted, that is true of most Class D under the shelf of a Class B. Even the other airports under the DFW Class B don’t have to do this. Addison is a very special situation due to its particular traffic. Though, any pilot approaching an uncontrolled airport this close to the surface area of a Bravo would benefit from local knowledge. As far as the DFW Bravo, in conversations with local TRACON, they would prefer all traffic (VFR included) to be on flight following or at least speaking to them when within the lateral boundaries of the Bravo’s outer ring. There’s no reason not to talk to approach


text:
What does setting an altimeter actually do to the altitude?
I have looked at probably 10 different webpages and re watched the Sportys video on altitudes, multiple times, but I have a very fundamental question, which I am unable to wrap my head around.
Is my following understanding correct?

Pressure altitude is the correction of altitude based on the air pressure on that day and time.
Density altitude is the correction of pressure altitude, based on the air temperature.

If the above statements are correct, what am I doing during this scenario:
The CFI asks me to listen to the weather at the airport, and get the altimeter reading. For example, the AWOS said 29.78. Now, I correct my altimeter in the aircraft based on that.
My dilemma here is... is this correction (from AWOS), only for pressure? - Meaning am I only correcting for pressure altitude?
Does this correction also involve temperature on that day? So my final altimeter setting is the density altitude.

Nothing happens to the altitude whatever you do with your instruments. What happens with the instrument reading is the real question...

Standard altimeters do not display altitudes corrected for non ISA temperature. There is such a thing as a pressure-temperature-corrected altitude, but it is essentially an engineering parameter (invisible to the pilot) only relevant in the context of RNP-AR approaches.

Fundamental is the understanding of "pressure" as a function not only of molecules per volume of air, but also how fast they move (temperature), and the content of the air (humidity).  John K's answer provides a great discussion of not only higher than standard Temps, but also lower ones.  The other way of doing it is to radio your destination (of known altitude) and ask for the altimeter barometric setting that yields that altitude, which will be 0 AGL for your landing.

Pressure Altitude is the height above sea level in the Standard Atmosphere (barometric pressure 29.92 in/hg, 1013 Mb). It's useful to note that absolute air pressure pressure at sea level in the Standard Atmosphere is 14.7 psia (absolute pounds per square inch).
Since the altimeter is just an absolute air pressure gauge (comparing ambient static pressure to a vacuum), this means that when the baro setting window is at 29.92, a pressure altitude indication of zero feet means the ambient static pressure at the altimeter is 14.7 psia.
Obviously, the absolute air pressure drops as you go up and increases as you go down; at 1000 ft above sea level on a standard day absolute air pressure is 14.2 psia, and if you went into a mine 1000 ft below sea level, the absolute air pressure at 1000 ft below sea level is 15.2 at a standard barometric pressure of 29.92/1013 (based on the standard atmosphere's pressure profile and lapse rate).
So think of pressure altitude as "height above/below the vertical point at which absolute pressure is 14.7 psia" wherever that happens to be, vertically speaking.
When high pressure (atmosphere is thicker where you are) or low pressure (atmosphere is thinner) weather systems are present, the level at which absolute air pressure is 14.7 psia will be above or below sea level.  If a high pressure system is present, say with 30.92" of mercury, the absolute pressure at sea level will be 15.2 psia, not 14.7.  If the altimeter is still at at 29.92, 15.2 psi absolute means the altimeter thinks it's at 1000 ft below sea level, and will show minus 1000 ft and you'll have to climb to 1000 ft true altitude for the altimeter to show zero.
Say the local altimeter setting is 30.92". When you turn the altimeter calibration knob to 30.92 in the window, and you are at sea level, you are adjusting the clock mechanism to make the altitude read zero feet at 15.2 psia, not 14.7 psia, to compensate for the higher pressure.  Or if you are at 1000 feet elevation, where the standard absolute pressure is 14.2, you are setting the altimeter to read 1000 ft above sea level at an absolute pressure of 14.7 psia.
So you are correct about the pressure setting; there is no temperature correction for density in the altimeter setting provided.  It's based on the standard atmospheric temperature of 59F/15C at sea level and the standard lapse rate going up.  Due to density effects, temperature deviations from standard affects the pressure being sensed by the altimeter's aneroid chamber in a similar way to pressure deviations from standard; that is, temperature below standard has the same effect as pressure above standard, and vice versa.  So when temperature is below standard, say it's near freezing, the altitude reading is higher than true altitude even when set to the correct barometric pressure, and vice versa when temperature is above standard.
However, these altitude deviations from temperature are not enough to bother with and we normally live with the error (everybody in your vicinity is following the same error, so no harm no foul from that aspect) except where the deviation becomes potentially dangerous. Potentially dangerous means the error is enough to eat into altitude margins when close to the ground or obstacles, and you can't see the ground or obstacles.  When it gets really cold, the errors can get quite large, even larger the farther above above the barometric pressure source the altitude we are concerned with is.
This gets dangerous when flying IFR approaches in winter, where a 100 ft error can be lethal when precise obstacle clearance observance is critical. So when using level-off altitudes within an IFR approach procedure, and the ambient temperatures are generally at or below freezing, you have to consult a temperature correction chart and apply correction values to level off altitudes and decision altitudes within the approach.

Outside of that circumstance (and sometimes enroute altitudes, if cruising close to minimum enroute altitudes in extreme cold where obstacle clearance is a concern), we don't normally bother with non-standard temperature corrections to altimeter readings, only the effect of high temperature related density reductions on takeoff performance.

Could you provide a better image of that cold temperature chart or a source pdf to look at it more clearly? Your chose to make the image the lowest resolution possible actually made it too low to clearly read.

AIM reference:  https://www.faa.gov/air_traffic/publications/atpubs/aim_html/chap7_section_3.html
Higher-res image:  https://www.faa.gov/air_traffic/publications/atpubs/aim_html/images/aim_img_33fcf.png


text:
What happens when I set my altimeter at the starting of my flight?
When I set my altimeter to 29.92 mm Hg, I have my pressure altitude.
However, what am I doing, when I set my altimeter before each flight. My instructor tells me to hear for AWOS, and set the altimeter accordingly.
Lets say I set it up to 29.77 mm Hg, what is the value I have on my altimeter now?
It is not pressure altitude. My instructor also mentioned the AWOS at the airport, is not corrected for Temperature.

Can you clarify exactly what you need to know that wasn't in your earlier question about altimeter settings? You accepted an answer that may cover some of this already. We also have this question that might help.

I very much doubt you set your altimeter to 29.77 mm Hg.....

@757 I don't doubt the setting, I doubt what OP thinks the units are.

When you set the local setting in the Baro window of the altimeter, it should indicate your actual field elevation.
It works the other way around.  If you don't have an altimeter setting, you dial the knob on the altimeter until it shows your field elevation.  Voila!  You're local altimeter setting is displayed in the Baro window.
Temperature corrections are not applied to altimeter indications except when it's below freezing and you are flying an IFR approach, where the errors can cut into obstacle clearance margins depending on how cold and how far above the setting source you are.  The colder, and the farther above the setting source, the bigger the error.
The rest of the time, temperature errors from being above or below standard (59F/15C at sea level) are no big deal because they are small and everybody around you has the same error, so it's not worth the trouble to apply a correction for a few feet here or there.

Op also seems to want to know that setting QNH shows his true altitude, rather than pressure altitude or density altitude, at least while he's on the ground.

Oh, I see the other answer answered that.

Field elevation is true altitude.  Same thing anyway.


text:
Overheating (fire thrown out of the nozzle)
At first I copied https://www.jetcat.de/en/productdetails/produkte/jetcat/produkte/hobby/Engines/P160_rxi_b
Where I manufactured and installed all the parts and they are very similar to the original product (also similar to the same quality of metal)
I am facing two problems:
The first: When the engine is running, the engine does not start properly (taking into account that the balance is not good because it contains vibration) and also makes a popping sound.

The second problem: Since the start of operation, the temperature begins to rise dramatically, which causes the turbine disc and the nozzle to redden, as it throws a large proportion of the fire outside the combustion chamber.
Notes:
The work circuit is manufactured and programmed manually by me:
The kerosene pumping rate works with a continuous mechanism (I imitated the pumping algorithm in the main circuit) and I am not sure if I connected it correctly. After searching, I found that the fuel pumping algorithm should be used in particular, but I did not find sources to help me.
Are there any suggestions to help me find solutions to my problems?

Are you asking engineering suggestions based on an engine picture? That said this is an impressive achievement.

Amazing project! Bravo!

As it is, the way this question is asked is asking for opinions and much much too broad. Pick one of your issues (engine does not start properly, engine is not balanced, non-stable temperature, uncontrolled (or not properly controlled) fuel flow) and give us as much information as you are willing and ask a concrete and answerable with facts question on that single point. (and oh by the way, well done project, so far)

I’m voting to close this question because it about home made version of a MODEL aircraft engine and has very little in common with actual aircraft engines.

We have a SE just for that - http://drones.stackexchange.com/ - this would be a good question there, with more detail though.

It sorta looks like you are burning all the air going through the combustion can.  Normal jet engines only burn a minor fraction of the through flow and use the rest to cool the engine and contain the flame within the combustion can.  Is that what the Jet Cat does?

@JohnK Yes I have this problem? Can you guide me to the solution?

Well you'd have to have holes around the burner can and duct some of the air from the compressor around it on the inside and outside walls so that the air is forced into the can through the holes, which is how normal engines do it.  The details? I have no idea. Get a book on jet engine theory and start experimenting.


text:
Unclear about how you can have high RPM but no power
I did see this question: Manifold pressure vs power vs rpm
But I'm still a bit unclear about the following. I've only flown cherokee pa28s, so I might not have empirical knowledge of the following.
I have a throttle lever in the plane which if I push up, the RPM goes up and it seems like propeller spins faster and the plane has more power, like when we climb out during takeoff.
But I was recently watching this video about Bonanza 6861Q which crashed. He said he had full manifold pressure and full RPM but no power. What does that even mean?

In the archer, I've never heard any of my CFIs talk about the manifold pressure. We check oil pressure, ammeter, and the vacuum gauge. Does manifold pressure refer to one of those things? If not, then what is the manifold pressure?
The RPM and power seem synonymous when I fly in my archer. I push up on the throttle, RPM goes up, feels like the plane has more power. Is RPM different from power?
If so, does the throttle increase the power or the RPM?

If your prop falls off you’ll get higher than usual RPM and no power! ;-)

@MD88Fan: Proof. ;-)

When you say "power" do you mean "thrust"?

@CaiusJard, well, if you don't have power, you can't have thrust either, but with propeller engines usually power, not thrust, is the characteristic variable we talk about. And power delivered to the propeller at that, not all of which gets converted to thrust.

The PA-28 Archer uses a fixed pitch propeller. Typically an aircraft with such a propeller, they are not fitted with a manifold pressure gauge as a means for gauging power output. Power output on a fixed pitch propeller is gauged by the engine tachometer. A Beech  Bonanza is equipped with a constant speed propeller and will also be equipped with a manifold pressure gauge. This gauge will be used as a measure of power output for the engine.
You really won’t get into the use of manifold pressure until you start flying complex airplanes which have a controllable pitch propeller.
It’s not that a PA-28 engine does not have manifold pressure associated with it. It does. But there is no gauge mounted in the aircraft to measure it.
Neither manifold pressure, nor engine speed are exacting indications of power output; they are sloppy indications and can just be used as a general yardstick to approximate how much power an engine is producing, similar to the torque gauge in a turbopropeller airplane or an N1 or EPR gauges for measuring thrust output from a jet.  In fact if you look at the cruise performance charts for a complex airplane, you’ll know that a particular power output requires both a manifold pressure and propeller speed setting in order to ensure an exact power output from the engine.  Similarly in an airplane with a fixed pitch propeller they will do a general gauging of power based on propeller speed in the cruise performance charts.  Note this power varies with altitude i.e. you will not get the same power out of the engine at 2400 RPM while flying at 6000 feet, versus flying at 3000 feet.
Remember that in an engine manifold pressure describes in fact the pressure of the fuel air mixture entering the engine manifold. When the engine is not operating, the manifold pressure will return to whatever the ambient atmospheric pressure is. Should an engine failure occur, the manifold pressure will not change, and will remain constant with a throttle setting provided the engine continues to turn.   Typically manifold pressure is measured in inches of mercury, so for that bonanza if it’s sitting on the ground, engine off, at sea level @ STP, the manifold pressure gauge would read 29.92 inches of mercury.
Neither manifold pressure nor engine RPM are a reliable indication of an engine failure and should never be used as such for the reasons stated above. That was the problem the pilot was perceiving when he lost engine power. As soon as the engine power went the manifold pressure gauge returned to ambient atmospheric pressure, which coincidentally is exactly the same as if the engine was operating with a throttle wide open. Propellers can often windmill at maximum speed in a dive, which also prevents them from being a reliable indication of engine power.   Your two most reliable gauges for determining engine failure in an AvGas powered piston airplane are 1) the fuel flow gauge and 2) the exhaust gas temperature (EGT) gauge or turbine inlet temperature gauge (TIT) for a turbocharged airplane.  That seems counterintuitive, but for the problems I listed above with manifold pressure gauges it’s the only reliable way to determine whether engine is properly operating.
The throttle in a light training airplane like a PA-28 only controls the butterfly throttle valve in a carbureted airplane, or the fuel/air control unit in a fuel injected airplane.  Engine tachometer only measures the rotational speed of the driveshaft and propeller

“As soon as the engine power went the manifold pressure gauge returned to ambient atmospheric pressure” – no, the manifold pressure really remained unchanged. Manifold pressure depends on how much air the engine sucks in and the throttle position only, and with the prop governor keeping constant RPM the engine is still pulling through the same amount of air when driven by the windmilling propeller. Only on a turbocharged engine the MP would (gradually) drop as the turbo winds down, because the exhaust is no longer expanded through combustion.

Also if you do have a torque monitor that is an accurate indicator, because power equals torque times angular velocity (RPM), by definition. So if you have those two, you know the power, exactly. For that reason, auto-feather usually uses negative reading on torque as trigger. But I don't think any light piston, even complex, has torque indicator. Only turbines, and some vintage large pistons (like DC-6 or Constellation) do.

… you can also still have fuel flow and no power if it's ignition that gave out, so that leaves EGT as the only reliable indicator the engine is not working properly.

… if the engine actually stopped, the manifold pressure would return to ambient, but in this case it didn't stop, it was still windmilling at the constant RPM.

Well the same could be set for a piston airplane in that if you have a manifold pressure gauge plus a engine tachometer, you can gauge how much power you are producing based on a conversion table. In addition your comment about toric is an accurate, as the displayed engine speed would have to be in rads per second to work without a conversion factor of some sort.

No, you can't, and that's exactly what this accident, and disussion demonstrates. On a normally-aspirated engine when you turn off the ignition, but let the governor keep the RPM of the widmilling propeller constant, neither MP, RPM or fuel flow will move but for a tiny fluctuation as the governor catches up.

As for units, it does not matter what units you have. If you have torque in $\mathrm{lb_f}·\mathrm{ft}$ and angular speed in revolutions per minute, you get power in $π \frac{\mathrm{lb_f}·\mathrm{ft}}{\mathrm{min}}$, which is somewhat weird unit, but it's just a unit and it is still a unit of power (multiply by 1 expressed as $0.071 \frac{\mathrm W}{π \frac{\mathrm{lb_f}·\mathrm{ft}}{\mathrm{min}}}$ or $9.5·10^{-5} \frac{\mathrm{hp}}{π \frac{\mathrm{lb_f}·\mathrm{ft}}{\mathrm{min}}}$ to get something reasonable).

I don’t think you’re understanding what I’m saying in my comments Jan. I’m not talking about the event of an engine failure

That still remains sloppy, Jan. Well torque and power use the same units, a torque gauge is in fact not indicating your engines power output, and you will have to perform the additional calculations you suggest above. Again it can serve as a general yardstick to use for particular flight situations i.e. on approach at Vref, a/c configured for landing and trimmed neutral the airplane remains stabilized at 500 ft-lbs, max RPM, etc.

The whole question is about engine failure, and the difference in response to engine failure was the key point of my comment that you responded to. 2. It isn't sloppy. Torque times angular velocity is power. If you have those two gauges – I never said torque alone – you know the power, and you know it as long as the gauges themselves, independent of anything else, work. Unlike MP, which needs many further assumptions. 3. No, torque and power definitely don't have the same unit, nor do they have the same dimension.

"a beach Bonanza" - would it need a sand filter?

From a physics standpoint, power is defined to be (force x velocity); in rotary shaft work terms this is (torque x RPM).
This means that a device that produces 100 ft-lbs of torque at zero RPM (shaft-locked condition) is producing zero power, and a device that produces 2500RPM with the load disconnected (no-load condition) is producing zero power.
If that device produces 100 ft-lbs of torque at 2500RPM, then it is producing (100 x 2500/5252) shaft horsepower where the 5252 divisor converts the units to horsepower.
With a fixed pitch propeller, the relationship between the engine load and the prop RPM is fixed, which means if you know your RPM, you can look up the corresponding horsepower on a chart displaying a power vs. RPM line.
With a variable-pitch propeller, that same chart will contain not one line but a family of lines corresponding to different values of the load, as determined by different values of the propeller pitch- and in this case each line is labelled by the manifold vacuum which has been measured on the engine running at that load level and RPM.
Note that manifold vacuum is a proxy variable for the engine load torque since it is a measure of how fast the engine is pulling fuel/air mixture through the manifold.

In rotational motion, HP = Ft-Lbs * RPM / 5252. The 550 is for converting HP = Ft-Lbs/s / 550, or linear motion. (You could also do radians per second or something and get other constants.)

I will edit! -NN

As already mentioned, a Beechcraft Bonanza has a constant speed propeller.

Constant speed propeller has a governor that drives the propeller coarser if it's too fast and finer if it's too slow. When the engine loses power, the propeller is driven to fine where it can maintain the RPM just through windmilling so RPM won't change.
Manifold pressure indicates how the pressure is reduced by the throttle body limiting the amount of air being inducted in the engine. Since at the same RPM the engine still inducts the same volume per second and the throttle body didn't move, the manifold pressure does not move either. It would only move if the problem was inlet obstruction.
In this case, the fuel flow could have been seen dropped as the cause was fuel line clogging, but if it was ignition failure, the fuel would still be inducted too.
Only EGT, exhaust gas temperature, dropping is a reliable indication that there no (or less) combustion going on and therefore the engine is failing. And the feel, of course.

If it was a turbocharged engine, the manifold pressure would drop (with some delay) as the turbo would wind down, because without combustion the volume of exhaust is less and wouldn't be able to drive it any more. But with normally aspirated engine it usually just stays the same in a failure. It is a useful reference for accurately setting the standard cruise power, and to avoid overboosting a turbocharged engine on take-off, but does not help in case of failures.


text:
Fixed-Wing Stall Speed Equation Valid for Differing Planetary Conditions?
For some time I have been looking for a simple stall speed equation that can factor in differing planetary conditions (gravity and air density) and changing vehicle mass, but have failed to find anything further than the standard stall speed equation:
v_stall = sqrt(2 * W/rho * S * C_L)
where,
v_stall = current stall speed
W = weight,
rho = air density,
S = wing area,
and C_L = max lift coefficient.
Additionally, I was looking for an equation that did not rely on wing area or lift coefficient. So, I derived my own expression. To be honest, I do not really remember how I came to it, I have been using it the past few years for hobby and conceptual design stuff, but it is the following:
v_stall = v_stall, 1g * sqrt( (1.225/rho_current) * (g_current/9.81) * (m_current/m_original) )
where,
v_stall = current stall speed
v_stall, 1g = stall speed under standard Earth conditions,
rho_ current = current air density,
g_current = current gravitational acceleration,
m_current = current vehicle mass,
m_original = original vehicle mass
It must be noted that v_stall, 1g is vehicle stall speed under 1 g and 1.225 kg/m^3 (standard Earth conditions)
This equation does work under theory and delivers results within a 1% to 3% difference from the standard equation. I ran calculations under many different air densities and gravitational accelerations, and the results almost always come startlingly close to the standard equation.
Can anyone else verify the validity of this equation?
Sorry for the poor formatting. I am new here.

You are correct. The maximum coefficient of lift (CL_max) is non-dimensional, so it will be the same regardless of gravity, density, speed, mass etc. (it is not perfectly non-dimensionalized, as flow regime does still vary with Reynolds number, Mach number). Because CL_max doesn't change, we can say:
CL_max1 = CL_max2

Taking the stall equation:
CL_max = W / (0.5 * rho * v_stall^2 * S)

We can substitute it into the first equation (twice):
W1 / (0.5 * rho1 * v_stall1^2 * S1) = W2 / (0.5 * rho2 * v_stall2^2 * S2)

Rearranging and simplifying:
 v_stall2 = v_stall1 * sqrt(rho1/rho2 * S1/S2 * W2/W1)

If we assume that the area is unchanged, and we represent W2/W1 as mass*gravity, then this comes out to the same as your equation:
 v_stall2 = v_stall1 * sqrt(rho1/rho2 * g2/g1 * m2/m1)

These relationships hold true so long as we assume an identical Reynolds & Mach numbers. As you are looking at different planets, make sure your Knudsen number remains < 1e-4 too, otherwise, the atmosphere is too rarefied for this relationship to hold.

Wow, this is an awesome answer. Thank you for deriving it!

Thank you. Please mark it as the answer and upvote.

Stall speeds vary widely based on many different factors, some of which you have listed already. However, an airfoil will always stall when it exceeds the critical angle of attack -- the point at which the CL or coefficient of lift reaches its maximum -- regardless of the configuration of the aircraft at any given moment. That is why pilots often fly to an AOA indicator first, and other factors second. Because of the way AOA works, air density is not relevant. Neither is gravity. The sensors (of various types) are used to calculate the angle of attack and tell you how close you are to stalling the wing. You may want to check out this link for a concise but very useful explanation of how AOA indicators work.

Yes, basically you just calculate a new stall speed for an airplane where mass, air density and gravity change, but wing area and maximum coefficient of lift remain constant. It is valid if similar airspeeds result in similar flow behavior (Reynolds number is the same), so the maximum coefficient of lift is actually the same. This may not be true if the atmosphere is made up of different gases, at a different temperature, or it is of such low density that you have to fly much faster and encounter compressibility effects. In this case the critical angle of attack also changes.

This is true, I have noticed nonsensical results for very low air densities and unreasonable temperatures.


text:
Can anyone identify this Biplane flown Army Air Reserve stationed at Clover Field, Santa Monica, California in mid the 1920's?
Can anyone identify this biplane flown by the 476th pursuit squadron "The Black Falcons" of the 322rd pursuit group of the Army Air Reserve stationed at Clover Field, Santa Monica, California in the mid to late 1920's? My grandfather was a member.

According to this PDF, their "Weapon System" was the PT-1, but the PT-1 at Wikipedia doesn't look like the plane in the picture.


text:
What biplane of the Army Air Reserve is in this picture?
This is the second example of the biplanes they flew.
Can anyone identify this biplane flown by the 476th pursuit squadron "The Black Falcons" of the 322rd pursuit group of the Army Air Reserve stationed at Clover Field, Santa Monica, California in the mid to late 1920's?

I did some looking could this biplane be a Consolidated PT-3?
The palm tree looking things is the Black Falcons Emblem not part of the aircraft. See this link. Third photo down https://thejivebombers.com/2018/01/23/486/
The struts on the left in the photo show a diagonal strut going to the rear of the upper wing. This is what made me think it might be a Consolidated PT.
I agree that the falcon is wearing a falconry hood. The ornament on top of the image is part of the hood. The hood covers the falcons eyes.

The one in the linked pic is a Jenny. This one isn't a Jenny, although it appears to have the same engine, an OX-5 V8.  The splayed cabane struts suggest a Waco product.

Hi Bill, it seems you created two separate accounts: this one and this one. See I accidentally created two accounts; how do I merge them? on how to merge them.

The engine looks like it has three cylinder banks (W engine), but the angles between them (45°) are unlike those of the Napier Lion.

Seems unusual that the falcon's fierce eye has not been depicted in the illustration (squadron logo). (Granted it's a silhouette, but...) Think there's any chance that the thing on top is a depiction of the ornament on a falconry hood (google it or see https://trc-leiden.nl/trc-needles/individual-textiles-and-textile-types/accessories/falcon-hood ), rather than a palm tree?  Hard to say, would seem odd for a military squadron to use a hooded bird on logo, but...